An organic compound undergoes first-order decomposition. The time taken for its decomposition to 1/8 and 1/10 of its initial concentration are \(\mathrm{t}_{1 / 8}\) and \(\mathrm{t}_{1 / 10}\) respectively. What is the value of \(\frac{[\mathrm{t}_{1 / 8}]}{[\mathrm{t}_{1 / 10}]} \times 10\) ? (take \(\log _{10} 2=0.3\) )

JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · Chemical Kinetics

An organic compound undergoes first-order decomposition. The time taken for its decomposition to 1/8 and 1/10 of its initial concentration are

t_{1/8}

and

t_{1/10}

respectively. What is the value of

\frac{[ t _{1/8} ]}{[ t _{1/10} ]} \times 10

? (take

lo g_{10} 2 = 0.3

)

Answer: 9

For a first-order reaction, the time for concentration to reach a fraction of its initial value is:

t = \frac{2.303}{k} lo g \frac{[ A ] _{0}}{[ A ]}

For

t_{1/8}

[A] = [A]_{0} /8

t_{1/8} = \frac{2.303}{k} lo g 8 = \frac{2.303}{k} \times 3 lo g 2 = \frac{2.303 \times 3 \times 0.3}{k}

For

t_{1/10}

[A] = [A]_{0} /10

t_{1/10} = \frac{2.303}{k} lo g 10 = \frac{2.303}{k}

Therefore:

\frac{t _{1/8}}{t _{1/10}} = \frac{2.303 \times 3 \times 0.3/ k}{2.303/ k} = 3 \times 0.3 = 0.9

\frac{t _{1/8}}{t _{1/10}} \times 10 = 0.9 \times 10 = 9

JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · Integer answer